Which descriptor applies to the circuit? A- on B- closed C- incomplete D- open

The displacement of a wave traveling in the positive x-direction is y(x, t)|= (3.5 cm)cos(2.7x − 92t), where x is in m and t is

zubka84 [21]

Answer

Given,

y(x, t) = (3.5 cm) cos(2.7 x − 92 t)

comparing the given equation with general equation

y(x,t) = A cos(k x - ω t)

A = 3.5 cm , k = 2.7 rad/m , ω = 92 rad/s

we know,

a) ω =2πf

f = 92/ 2π

f = 14.64 Hz

b) Wavelength of the wave

we now, k = 2π/λ

2π/λ = 2.7

λ = 2 π/2.7

λ = 2.33 m

c) Speed of wave

v = ν λ

v = 14.64 x 2.33

v = 34.11 m/s

5 0

3 years ago

Biologists use optical tweezers to manipulate micron-sized objects using a beam of light. In this technique, a laser beam is foc

vekshin1

Answer:

Explanation:

Part A) Using

light intensity I= P/A

A= Area= π (Radius)^2= π((0.67*10^-6m)/(2))^2= 1.12*10^-13 m^2

Radius= Diameter/2

P= power= 10*10^-3=0.01 W

light intensity I= 0.01/(1.12*10^-13)= 9*10^10 W/m^2

Part B) Using

I=c*ε*E^2/2

rearrange to solve for E= $\sqrt{$ ((I*2)/(c*ε))

c is the speed of light which is 3*10^8 m/s^2

ε=permittivity of free space or dielectric constant= 8.85* 10^-12 F⋅m−1

I= the already solved light intensity= 8.85*10^10 W/m^2

amplitude of the electric field E= $\sqrt{$ (9*10^10 W/m^2)*(2) / (3*10^8 m/s^2)*(8.85* 10^-12 F⋅m−1)

---> E= $\sqrt{$ (1.8*10^11) / (2.66*10^-3) = $\sqrt{$ (6.8*10^13) = 8.25*10^6 V/m

8 0

3 years ago

1 .

son4ous [18]

Answer:

Find answers below.

Explanation:

1. Radiant: the energy of light.

2. Light: makes it possible to see things.

3. Sun: a source of light. This source of light is referred to as solar energy and it's renewable source of energy.

4. Crest: top of wave. Thus, it's typically the highest part of an electromagnetic wave.

5. Trough: bottom of wave. Thus, it's typically the lowest part of an electromagnetic wave.

6. Wavelength: distance from one point on one wave to the same point on the next wave.

7. Infrared light: it has longer wavelengths than visible light.

8. X-rays: shorter wavelengths than visible light.

Electromagnetic waves is a propagating medium used in all communications device to transmit data (messages) from the device of the sender to the device of the receiver.

An electromagnetic spectrum refers to a range of frequency and wavelength that an electromagnetic wave is distributed or extends. The electromagnetic spectrum comprises of gamma rays, visible light, ultraviolet radiation, x-rays, radio waves, and infrared radiation.

3 0

2 years ago

For the long life cells we have to connect them in ____ combination

NISA [10]

Answer:

Parallel combination.

5 0

3 years ago

Read 2 more answers

The spring of a spring gun has force constant k = 400 N/m and negligible mass. The spring is compressed 6.00 cm and a ball with

nikdorinn [45]

Answer:

A) v = 6.93 m/s

B) v = 4.9 m/s

C) x_m = 0.015m

D) v_max = 5.2 m/s

Explanation:

We are given;

x = 6 cm = 0.06 m

k = 400 N

m = 0.03 kg

F = 6N

A) from work energy law, work dome by the spring on ball which now became a kinetic energy is;

Ws = K.E = ½kx²

Similarly, kinetic energy of ball is;

K.E = ½mv²

So, equating both equations, we have;

½kx² = ½mv²

Making v the subject gives;

v = √(kx²/m)

Plugging in the relevant values to give;

v = √((400 × 0.06²)/0.03)

v = √48

v = 6.93 m/s

B) If there is friction, the total work is;

Ws = ½kx² - - - (1)

Work of the ball is;

Wb = KE + Wf

So, Wb = ½mv² + fx - - - (2)

Combining both equations, we have;

½mv² + fx = ½kx²

Plugging in the relevant values, we have;

(½ × 0.03 × v²) + (6 × 0.06) = ½ × 400 × 0.06²

0.015v² + 0.36 = 0.72

0.015v² = 0.72 - 0.36

v² = 0.36/0.015

v = √24

v = 4.9 m/s

C) The speed is greatest where the acceleration stops i.e. where the net force on the ball is zero. (ie spring force matches 6.0N friction)

So, from F = Kx;

(x is measured into barrel from end where F = 0)

Thus; 6.0 = 400x

x_m = 6/400

x_m = 0.015m from the end after traveling 0.045m

D) Initial force on ball = (Kx - F) =

[(400 x 0.06) - 6.0] = 18N

Final force on ball = 0N

Mean Net force on ball = ½(18 + 0)

Mean met force, F_m = 9N

Net Work Done on ball = KE = 9N x 0.045m = 0.405 J

Thus;

½m(v_max)² = 0.405J

(v_max)² = 2 x 0.405/0.03

(v_max)² = 27

v(max) = √27

v_max = 5.2 m/s

6 0

2 years ago